Question

3x+2-2$\sqrt{x+3}$*$\sqrt{2x-1}$=$\sqrt{x+3}$-$\sqrt{2x-1}$

Answer 1

$3x+2-2\sqrt{x+3} \sqrt{2x-1}=x+3-2\sqrt{x+3} \sqrt{2x-1}+2x-1=(\sqrt{x+3})^{2}-2\sqrt{x+3} \sqrt{2x-1}+(\sqrt{2x-1})^{2}=(\sqrt{x+3}- \sqrt{2x-1})^{2}\\ y=\sqrt{x+3}- \sqrt{2x-1}\\\\y^{2}=y\\y_{1}=0, y_{2}=1\\y_{1})\\\sqrt{x+3}-\sqrt{2x-1}=0\\\sqrt{x+3}=\sqrt{2x-1}\\x+3=2x-1\\x_{1}=4\\y_{2})\\y_{2}^{2}=3x+2-2\sqrt{x+3} \sqrt{2x-1}=1\\2\sqrt{x+3} \sqrt{2x-1}=3x+1\\4(x+3)(2x-1)=(3x+1)^{2}\\8x^{2}+20x-12=9x^{2}+6x+1\\x^{2}-14x+13=0\\x_{2}=1, x_{3}=13$

Все три полученные ответа входят в ОДЗ ( x≥0.5)

Ответ: х∈{1; 4; 13}